Prove that if G is a group such that G/Z(G) is cyclic, then G is abelian. (I am unable to connect logic):furious:I don't know why?:confused::confused:

cjem

December 22nd, 2017 02:25 AM

Quote:

Originally Posted by shashank dwivedi
(Post 586214)

Prove that if G is a group such that G/Z(G) is cyclic, then G is abelian. (I am unable to connect logic):furious:I don't know why?:confused::confused:

Say the coset of $Z(G)$ represented by $g \in G$ generates $G/Z(G)$. Then every coset of $Z(G)$ is represented by $g^k$ for some $k$. Take any two elements $a, b$ of $G$. $a$ must be contained in a coset of $Z(G)$, say it's in the one represented by $g^n$. This means $a = g^n x$ for some $x \in Z(G)$. Similarly, $b = g^m y$ for some $m$ and some $y \in Z(G)$. Now it's straightforward to show that $a$ and $b$ commute.